Solve using Gaussian Elimination with back-substitution or Gauss Jordan 3z1 - 2x2 +4x3 1 21+2-2x3 = 3 %3D 221 – 3r2 +6x3 8 O 21 = 2, 22 = 4, 23 = 6 %3D %3D %3D O 21 = -3, 2 = -5, a3 = -7 %3D %3D O z1 = 3, 2 = 5, 23 7 %3D %3D %3D No solution
Solve using Gaussian Elimination with back-substitution or Gauss Jordan 3z1 - 2x2 +4x3 1 21+2-2x3 = 3 %3D 221 – 3r2 +6x3 8 O 21 = 2, 22 = 4, 23 = 6 %3D %3D %3D O 21 = -3, 2 = -5, a3 = -7 %3D %3D O z1 = 3, 2 = 5, 23 7 %3D %3D %3D No solution
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.1: Systems Of Equations
Problem 6E
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Question
![Solve using Gaussian Elimination with back-substitution or Gauss Jordan
3z1 – 2x2 +4x3 = 1
2i + x2 - 2x3 = 3
%3D
221 – 3z2 + 6x3 = 8
%3D
O 21 = 2, 22 = 4, 13 = 6
%3D
%3D
O 21 = -3, 2 = -5, 23 = -7
%3D
%3D
21 = 3, 22 = 5, 23 = 7
%3D
%3D
%3D
No solution](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb28b4845-2cf8-47d5-be9f-7cfb4b1b7d89%2F0edff8ce-85a7-4809-8eb0-5f3673e2b244%2Fspd7mr7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solve using Gaussian Elimination with back-substitution or Gauss Jordan
3z1 – 2x2 +4x3 = 1
2i + x2 - 2x3 = 3
%3D
221 – 3z2 + 6x3 = 8
%3D
O 21 = 2, 22 = 4, 13 = 6
%3D
%3D
O 21 = -3, 2 = -5, 23 = -7
%3D
%3D
21 = 3, 22 = 5, 23 = 7
%3D
%3D
%3D
No solution
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